\newproblem{lay:7_1_7}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.1.7}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Determine if the matrix $A=\begin{pmatrix} 0.6 & 0.8 \\ 0.8 & -0.6 \end{pmatrix}$ is orthogonal. If it is, find its inverse.
}{
   % Solution
	A matrix $A$ is orthogonal if all its columns are orthogonal to each other and they are of unit norm. In this case
	\begin{center}
		$\left<(0.6,0.8),(0.8,-0.6)\right>=0.6\cdot 0.8 + 0.8\cdot (-0.6)=0$ \\
		$\|(0.6,0.8)\|^2=0.6\cdot 0.6 + 0.8\cdot 0.8=1$ \\
		$\|(0.8,-0.6)\|^2=0.8\cdot 0.8 + (-0.6)\cdot (-0.6)=1$ \\
	\end{center}
	Since the two columns are orthogonal to each other, $A$ is an orthogonal matrix. The inverse of an orthogonal matrix is its transpose. In this case
	\begin{center}
		$A^{-1}=A^T=\begin{pmatrix} 0.6 & 0.8 \\ 0.8 & -0.6 \end{pmatrix}$
	\end{center}
}
\useproblem{lay:7_1_7}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

